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2 years ago

An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?

Physics

1 answer:

Serjik [45]2 years ago

6 0

The image of the object is 8cm to the left of the lens (D)

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

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$\Delta U = Q-W$

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Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

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Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$